A group has $14$ men and $11$ women.
(a) How many different teams can be made with $7$ people?
(b) How many different teams can be made with $6$ people that contains exactly $4$ women?

Answer key to a is $257$ but I can't figure out how to get $257$? There's no answer key to b though, but here's my attempt:

$$\binom{25}{6} - \left[\binom{14}{6} + \binom{14}{5} + \binom{14}{4} + \binom{14}{3} + \binom{14}{1} + \binom{11}{6} + \binom{11}{5}\right]$$

What I'm trying to do here is subtracting all men, all women, 5 men, 4 men, 3 men, 1 men, and 5 women team from all possible combination of team.


  • $\begingroup$ You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. $\endgroup$ – Zev Chonoles Apr 8 '13 at 9:59
  • $\begingroup$ Are you sure the answer key to (a) is not \binom {25} 7 i.e. $\binom {25} 7$? $\endgroup$ – Lord_Farin Apr 8 '13 at 10:08
  • $\begingroup$ Ya that's what confuses me. It's just 257 and another similar problem has a similar small number when my answer is something bigger such as this. Maybe typos? Otherwise, it's pretty straight forward. $\endgroup$ – Andy Apr 8 '13 at 10:12


For the first one, number of ways you can choose $k$ woman and $m$ men= $11\choose k$+ $14\choose m$, such that $k+m=7$.

For the second one, number ways you can choose $4$ women= $14\choose4$

AND number of choosing $2$ men= $11 \choose 2$, when you have an AND, you gotta multiply them.

  • $\begingroup$ For the second, you swapped the number of men and women. $\endgroup$ – gnasher729 May 1 '14 at 23:45

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